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Problem Statement
Given a set of positive numbers, determine if a subset exists whose sum is equal to a given number ‘S’.
Example 1:
Input: {1, 2, 3, 7}, S=6
Output: True
The given set has a subset whose sum is '6': {1, 2, 3}
Example 2:
Input: {1, 2, 7, 1, 5}, S=10
Output: True
The given set has a subset whose sum is '10': {1, 2, 7}
Example 3:
Input: {1, 3, 4, 8}, S=6
Output: False
The given set does not have any subset whose sum is equal to '6'.
Constraints:
1 <= num.length <= 200
1 <= num[i] <= 100
Basic Solution
This problem follows the 0/1 Knapsack pattern and is quite similar to Equal Subset Sum Partition
. A basic brute-force solution could be to try all subsets of the given numbers to see if any set has a sum equal to ‘S’.
So our brute-force algorithm will look like:
for each number 'i' create a new set which INCLUDES number 'i' if it does not exceed 'S', and recursively process the remaining numbers create a new set WITHOUT number 'i', and recursively process the remaining numbers return true if any of the above two sets has a sum equal to 'S', otherwise return false
Since this problem is quite similar to Equal Subset Sum Partition
, let’s jump directly to the bottom-up dynamic programming solution.
Bottom-up Dynamic Programming
We’ll try to find if we can make all possible sums with every subset to populate the array dp[TotalNumbers][S+1]
.
For every possible sum ‘s’ (where 0 <= s <= S), we have two options:
- Exclude the number. In this case, we will see if we can get the sum ‘s’ from the subset excluding this number =>
dp[index-1][s]
- Include the number if its value is not more than ‘s’. In this case, we will see if we can find a subset to get the remaining sum =>
dp[index-1][s-num[index]]
If either of the above two scenarios returns true, we can find a subset with a sum equal to ‘s’.
Let’s draw this visually, with the example input {1, 2, 3, 7}, and start with our base case of size zero:
Code
Here is the code for our bottom-up dynamic programming approach:
Time and Space Complexity
The above solution has the time and space complexity of O(N*S), where ‘N’ represents total numbers and ‘S’ is the required sum.
Challenge
Can we improve our bottom-up DP solution even further? Can you find an algorithm that has O(S) space complexity?
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